Сподробным решением если можно 1) (√2-√2i)^4 2) (-1-√3i)^8 3) (√3+2√3i)^2

1) (√2-√2i)^4=[(√2-√2i)²]²=(2-4i+2i²)²=(4-4i-2)²=(2-4i)²=4-16i+16i²=
=4-16i-16=-12-16i
2) (-1-√3i)^8=[(-1-√3i)²]^4=(1+2√3i+3i²)^4=(1+2√3i-3)^4=(-2+2√3i)^4=
[(-2+2√3i)²]²=(4-8√3i+12i²)²=(4-8√3i-12)²=(-8-8√3i)²=64+128√3i+192i²=
=64+128√3i-192=-128+128√3i
3) (√3+2√3i)^2=3+12i+12i²=3+12i-12=-9+12i